I' m a worker and I'm self studying stochastic calculus. I cannot find a proof of the statement in the title; everybody seems to take it as trivial.
Consider two measurable spaces $(\Omega, \mathcal{F})$ and $(E, \mathcal{E})$, where $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$ and $\mathcal{E}$ is a $\sigma$-algebra on $E$. Now, if we take a discrete time process: $X=\{X_n \}_{n \in \mathbb{Z}^+ }$; I wish to show that the map
$$
(n,\omega)\mapsto X_n(\omega)
$$
is measurable with respect to the product $\sigma$-algebra $\mathcal{T}\otimes \mathcal{F}$, where $\mathcal{T}$ is a $\sigma$-algebra on $\mathbb{Z}^+$ that we take as the power set of $\mathbb{Z^+}$.
By definition of stochastic process the map
$$
\omega\mapsto X_n(\omega)
$$
is always $\mathcal{F}$-measurable, and I know that the map:
$$
n \mapsto X_n(\omega)
$$
is $\mathcal{T}$-measurable since $\mathcal{T}$ contains all the subset of $\mathbb{Z}^+$.
From this I cannot succeed in prooving that for each $A \in \mathcal{E}$:
$$
\{ (n,\omega) : X_n(\omega)\in A \} \in \mathcal{T}\otimes \mathcal{F}.
$$
For the very same reason I cannot prove that and adapted stochastic process in discrete time is always progressively measurable.
Can you give me an hint or a reference?
Best Answer
Fix $A \in \mathcal{E}$. We have
$$\{(n,\omega); X_n(\omega) \in A\} = \bigcup_{n \in \mathbb{N}} \big( \{n\} \times \{\omega \in \Omega; X_n(\omega) \in A\} \big).$$
Since $$\underbrace{ \{n\}}_{\in \mathcal{T}} \times \underbrace{\{\omega \in \Omega; X_n(\omega) \in A\}}_{\in \mathcal{F}} \in \mathcal{T} \otimes \mathcal{F}$$
for each $n \in \mathbb{N}$, it follows that
$$\{(n,\omega); X_n(\omega) \in A\}$$
is a countable union of $\mathcal{T} \otimes \mathcal{F}$-measurable sets and, hence, $$\{(n,\omega); X_n(\omega) \in A\} \in \mathcal{T} \otimes \mathcal{F}.$$